Banach Space Theory: The Basis for Linear and Nonlinear by Marián Fabian, Petr Habala, Petr Hájek, Vicente Montesinos, PDF

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Suppose that, for some C > 0, we have x p2 ≤ C x p1 for all x ∈ X . Let 1 < p < p1 . Write p1 = 1 , λp + (1 − λ) p2 , 0 < λ < 1. By the Hölder inequality used for numbers λ1 and 1−λ we have 34 1 Basic Concepts in Banach Spaces x p1 p1 1 = 0 1 ≤ |x|λp · |x|(1−λ) p2 0 λ 1 (|x|λp ) λ 1 · 0 ≤ x 1 |x|λp+(1−λ) p2 = 1−λ 1 (|x|(1−λ) p2 ) 1−λ pλ p = x p2 (1−λ) p2 · x 0 pλ p · (C x p1 ) p2 (1−λ) = x · C p2 (1−λ) · x pλ p p2 (1−λ) p1 = x pλ p · C p2 (1−λ) · x p1 −λp . p1 Thus 1≤ x λp p · C p2 (1−λ) · x −λp p1 .

Thus T : F ⊕ F ⊥ → H defined by T (x, y) = x + y is an isomorphism of F ⊕ F ⊥ onto H , and so H is the topological direct sum of F and F ⊥. Proof: Let x ∈ H . We claim there is a closest element x1 to x in F and (x −x1 ) ⊥ F. Let yn ∈ F be such that x − yn 2 < d 2 + n1 , where d := dist(x, F) = inf{ x − z : 26 1 Basic Concepts in Banach Spaces z ∈ F}. We now prove that yn is a Cauchy sequence. 8): 2x − (yn + ym ) 2 + ym − ym 2 = 2 x − yn 2 + 2 x − ym 2 . We can estimate ym − ym − 2x − (yn + ym ) y n + ym 2 = 2 x − yn 2 + 2 x − ym 2 − 4 x − 2 ≤ 2 d 2 + n1 + 2 d 2 + m1 − 4d 2 = n2 + m2 , where we used Then x − x1 2 = 2 x − yn 2 + 2 x − ym 2 2 yn + ym ∈ F.

The sequence {(1, 1, . . , 1, 0, . . )}∞ n=1 is Cauchy and not convergent as the only candidate for the limit would be (1, 1, . . ) ∈ / c0 . 33 Let M be a dense (not necessarily countable) subset of a Banach space X . xk and x k ≤ Show that for every x ∈ X \{0} there are xk ∈ M such that x = 3 x . 2k Hint. Find x 1 ∈ M such that x − x1 ≤ x −(x1 +· · ·+ xk−1 )− x k ≤ x− k n=1 x n ≤ x 2k−1 + x 2k x 2k = x 2 . Then x = 3 x 2k , then by induction xk ∈ M such that xk and xk = x− k−1 n=1 x n − . 34 Show that a Banach space X is separable if and only if S X is separable.

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